Integral of Arcsin[x]

  1. I just can't see it. I would think that it would involve some form of trig substitution, but I'm just drawing a blank. I'll do the work if someone can please give me a nice little hint.

    I know that [tex]\int \arcsin{(x)}dx = \sqrt{1-x^2}+ x\arcsin{(x}) + C[/tex] from my calculator and mathematica.

    Hint please. :yuck:
     
  2. jcsd
  3. Hurkyl

    Hurkyl 16,090
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    If only you were working with its derivative! I bet you know how to integrate something that looks like its derivative!
     
  4. Sure... I'd draw a nice triangle and do trig substitution if it was [tex]\int \frac{dx}{\sqrt{1-x^2}}[/tex]

    Hmmm...I'll think out loud here. If the integral has a square-root and is in the form of [itex]c^2-x^2[/itex] then x is one leg, c is the hypotenuse, and the other leg is the previously mentioned radical.

    Is trig substitution with right triangles on the right track? Since the hypotenuse is [itex]\sqrt{a^2+b^2}[/itex], it seems that one leg might need to be [itex]\sqrt{\arcsin{(x)}}[/itex] Somehow I don't think I'm on track.
     
  5. Use integration by parts. Remember to let u=arcsin(x) and v=x.

    u sub: InverseLogAlgebraicTrigExp
     
  6. Hurkyl

    Hurkyl 16,090
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    Sheesh, just give him the answer, why don't ya? :grumpy:
     
  7. Ah, thank you! :surprised It's so simple.

    And thank you Hurkyl as well. I still have lots to learn.
     
  8. I apologize Hurkyl :smile:
     
  9. well, you know the integral of sinx with limits. Now arcsin x will be the limits, and you can make a rectangle.
     
  10. Or you could just take the derivative of the right hand side and go "ta da!" and that's proof enough for me.
     
  11. hint

    equate the arc sine to another variable e.g y.making it a sine fxn.e.g the arc sine of 0.5=30,while sine30=0.5.this will simplify the integral and further substitution will conclude it
     
  12. Tide

    Tide 3,143
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    Just for the fun of it ...

    The sum of the integrals

    [tex]\int \sin^{-1} x dx + \int \sin y dy[/tex]

    is just the area of the bounding rectangle: [itex] x \times \sin^{-1} x[/itex]

    Since [tex]\int \sin y dy = -\cos y + C[/tex] and [tex]\cos y = \cos \sin^{-1} x = \sqrt {1-x^2}[/tex] it follows that
    [tex]\int \sin^{-1} x dx = \sqrt {1-x^2} + x \sin^{-1} x + C[/tex]
     
  13. Integration by parts?

    Hey, Im really sorry to arise dead threads from the past (which i have seen though google) but somehting really weird happened me when I tried to use integration by parts on arcsinx.


    let me show you:
    S(arcsinx)= {v'(x)=1} {u(x)=arcsinx}

    xarcsinx-S(x*d(arcsinx))=

    xarcsinx-S(x/(1-x^2)^0.5= {u(x)=x v'(x)=arcsinx}

    xarcsinx-xarcsinx+S(arcsin)dx


    ==> S(arcsinx)=S(arcsinx)

    :\
    I know I have done something really stupid here, but please be easy on me since I started studying Integrals only three days ago.

    In the second time I used integration by parts, do I miss something , is there another efficient choice of v and u?

    Thanks in advance,
    Aviv

    p.s: I will edit it better to use normal math signs once I figure out how.
     
  14. Gib Z

    Gib Z 3,348
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    For the second time you integrate by parts, swap your choices.
     
  15. Im sorry, Tried it also and all i got is:

    xarcsine x -(x^2/2)(1/(Sqrt(1-x^2)))-(x/4)(sqrt(1/(1-x^2))+1/4(arcsinx)

    this isn't going anywhere :(
     
  16. dextercioby

    dextercioby 12,292
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    Make [itex] x=\sin t [/itex]. Then apply part integration on the resulting integral. It's just a way to avoid the simple solution of part integrating directly.
     
  17. Office_Shredder

    Office_Shredder 4,500
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    After the first integration by parts, I would use a substitution
     
  18. solve it officially. Fixed during some major mistakes I had about dev' and stuff.

    did it without subtition, only using integration by parts.

    if you are interested what I did then you are welcome to tell me to write my solution.


    Thanks guys :)
    gg
     
  19. use seperation by parts

    u=arcsinx
    du=dx/(1-x^2)^1/2

    dv=dx
    v=x

    uv-ingegral vdu = xarsinx-integral x/(1-x^2)^1/2
    use u substitution with u = 1-x^2 so du = -2x
    than you get

    xarsinx + (1-x^2)^1/2 + c
     
  20. f(x)=arcsinx f'(x)=1/radical(1-xsquare)
    g'(x)=1 g(x)=x

    // S means integral

    S arcsin x dx= x arcsin x - S xdx/radical(1-xsquare) = x arcsin x + S (radical(1-xsquare))'dx=
    =x arcsin x + radical(1-xsquare) +C
     
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